2D rotation and angular momentum uncertainty

[hokhani]

TL;DR

How a quantum particle may be confined to a planar rotation?

In quantum mechanics, the uncertainty among the components of the angular momentum doesn't allow to find all the components together. However, when a quantum particle is confined to the $xy-$plane, it seems that $L_x$ and $L_y$ are both zeros. How about the uncertainty here?

 

[pines-demon]

What do you mean? What do you want to know? In 2D you only have one generation of rotations: $L_z$.

 

[hilbert2]

Even in 3D there can be state vectors, such as the hydrogenic atom s orbitals, where all of the angular momentum components have value 0. The uncertainty relation just says that you can't form a complete basis of $\mathcal{H}$ from only vectors that have simultaneously definite $L_x$, $L_y$ and $L_z$.

 

[PeterDonis]

when a quantum particle is confined to the $xy-$plane, it seems that $L_x$ and $L_y$ are both zeros.

No, when a quantum particle is confined to the $x y$ plane, its Hilbert space is restricted, and there aren't even any such operators as $L_x$ and $L_y$ on vectors in the restricted Hilbert space (there is only one angular momentum operator on that space, $L_z$). So it doesn't even make sense to talk about any uncertainty relation between $L_x$ and $L_y$.

 

[pines-demon]

Even in 3D there can be state vectors, such as the hydrogenic atom s orbitals, where all of the angular momentum components have value 0. The uncertainty relation just says that you can't form a complete basis of $\mathcal{H}$ from only vectors that have simultaneously definite $L_x$, $L_y$ and $L_z$.

Why are we discussing the 3D case now? Anyway, in that case the uncertainty reads
$$\Delta L_x \Delta L_y \geq \frac12 |\langle [L_x,L_y]\rangle|=\frac\hbar 2 |\langle L_z \rangle |=\frac12 \hbar^2 |m_z| $$

as you chose an s-state we have that $\ell=0$ and thus $m_z=0$ so there is no issue with $\Delta L_x$ or $\Delta L_y$ being as small as 0.

 

[hilbert2]

Why are we discussing the 3D case now? Anyway, in that case the uncertainty reads
$$\Delta L_x \Delta L_y \geq \frac12 |\langle [L_x,L_y]\rangle|=\frac\hbar 2 |\langle L_z \rangle |=\frac12 \hbar^2 |m_z| $$

as you chose an s-state we have that $\ell=0$ and thus $m_z=0$ so there is no issue with $\Delta L_x$ or $\Delta L_y$ being as small as 0.

Just to point out that in individual cases, a vector $\left|\psi\right.\rangle$ can be a simultaneous eigenstate of non-commuting observables. If it's angular momentum in question, this can happen when eigenvalues are zero.